Problem: The matrix
\[\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}\]is its own inverse, for some real numbers $a$ and $d.$  Find the number of possible pairs $(a,d).$
Solution: Since $\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}$ is its own inverse,
\[\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}^2 = \begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix} \begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix} = \mathbf{I}.\]This gives us
\[\begin{pmatrix} a^2 - 24 & 3a + 3d \\ -8a - 8d & d^2 - 24 \end{pmatrix} = \mathbf{I}.\]Then $a^2 - 24 = 1,$ $3a + 3d = 0,$ $-8a - 8d = 0,$ and $d^2 - 24 = 1.$  Hence, $a + d = 0,$ $a^2 = 25,$ and $d^2 = 25.$  The possible pairs $(a,d)$ are then $(5,-5)$ and $(-5,5),$ giving us $\boxed{2}$ solutions.